Question

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Answer 1

Would it be a direct or indirect proof?

Answer 2

Hint :

\(x\mid y\) and \(x \mid z\) \(\implies \) \(x\mid (y-z)\)

Answer 3

hmmm

Answer 4

\( b \mid a\) and \(b \mid (a + 2)\) \(\implies \) \(b\mid 2\)


What can you infer from above ?

Answer 5

im confused because of the fact that a has to be a positive integer

Answer 6

I just don't see it...

Answer 7

positive or negative, it doesn't matter

Answer 8

well i see how b=2 but how does it equal 1?

Answer 9

Look at my second reply from top.
Do you get why \(b\mid 2\) ?

Answer 10

nuh uh

Answer 11

Do you understand my first hint ?

Answer 12

yes, so you're just moving it around?

Answer 13

see if u can give me an example of that hint

Answer 14

You just need to understand that hint for the proof

Answer 15

if 5|20 and 5|10 then 5|(20-10)?

Answer 16

but I don't get how you know that

Answer 17

Good. Lets prove the result in that hint first

Answer 18

well 5|5

Answer 19

sorry 5|10

Answer 20

I've attached a proof of @ganeshie8 's claim that he made in the hint. I'm doing it as text document so I don't barge in too much into the conversation

Answer 21

that still has me a little lost

Answer 22

which line are you stuck at ?

Answer 23

the subtraction

Answer 24

(of the two equations)

Answer 25

why not factor out the gcf beforehand?

Answer 26

I used the idea that if `A = B` and `C = D` then you can subtract the two equations (left sides grouped together, same with the right)

so we will end up with `A - C = B - D`

notice how the order stays the same. A comes first and so does B

Answer 27

ohh alright

Answer 28

that makes sense

Answer 29

so we assume b|a and b|(a+2) then b is a common factor of (a, a+2) by the propstition of divisibility?

Answer 30

so b divides any linear combination of (a, a + 2)

Answer 31

therefore means b|[(a+2)- a] i.e b|2

Answer 32

But is that a valid proof?

Answer 33

the proposition of divisibility is used was the ( if a|b and a|c then a|(bx+cy) )

Answer 34

` we assume b|a and b|(a+2) then b is a common factor of (a, a+2)` I agree

if x | y and x | z then x is a common factor of y and z

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`b divides any linear combination of (a, a + 2)` I also agree here too

the linear combination that is handy here is `-1*a + 1*(a+2)` which ultimately simplifies to 2

Answer 35

So I guess it's right, well I don't see anything wrong with it. Am I missing anything?

Answer 36

Your concern is valid. This would a valid proof only if you had proven below previously :

`if a|b and a|c then a|(bx+cy)`

Answer 37

well that's one of our notes, we don't need to re-prove something we did in class, just refer to it (prof's decision)

Answer 38

I know technically I should prove it, I guess I can re-write the proof if I can find it.

Answer 39

keep in mind that `x-y` is a linear combination of x and y

it is in the form `bx+cy` where b = 1 and c = -1

so the two ideas are very similar

Answer 40

I think that method is just fine, the a | (bx+cy) method

Answer 41

awesome, thanks guys you're both amazing <3 haha, a lot smarter than me for sure :P

Answer 42

\(b\mid a\) and \(b\mid(a+2)\) \(\implies \) \(b \mid 2\)

What can you conclude from \(b\mid 2\) ?

Answer 43

if you're curious about the more generalized linear combination proof, see attached

Answer 44

Thanks!